a) Ta có $P(x)-Q(x)=(x^3-3x^2+x+1)-(2x^3-x^2+3x-4)$

$=x^3-3x^2+x+1-2x^3+x^2-3x+4$

$=-x^3-2x^2-2x+5$.

b) Thay $x=1$ vào hai đa thức ta có:

$P(1)=1^3-3.1^2+1+1=0$

$Q(1)=2.1^3-1^2+3.1-4=0$

Vậy $x=1$ là nghiệm của cả hai đa thức $P(x)$ và $Q(x)$.

a) Ta có $P(x)-Q(x)=(x^3-3x^2+x+1)-(2x^3-x^2+3x-4)$

$=x^3-3x^2+x+1-2x^3+x^2-3x+4$

$=-x^3-2x^2-2x+5$.

b) Thay $x=1$ vào hai đa thức ta có:

$P(1)=1^3-3.1^2+1+1=0$

$Q(1)=2.1^3-1^2+3.1-4=0$

Vậy $x=1$ là nghiệm của cả hai đa thức $P(x)$ và $Q(x)$.

a)\(\dfrac{x}{-4}=\dfrac{-11}{2}\)

x*2=-11*(-4)

x*2=44

x=44/2

x=22

b)\(\dfrac{15-x}{x+9}=\dfrac{3}{5}\)

(15-x)*5=(x+9)*3

$75-5x=3x+27$

$8x=48$

$x=6$.

a)\(\dfrac{x}{-4}=\dfrac{-11}{2}\)

x*2=-11*(-4)

x*2=44

x=44/2

x=22

b)\(\dfrac{15-x}{x+9}=\dfrac{3}{5}\)

(15-x)*5=(x+9)*3

$75-5x=3x+27$

$8x=48$

$x=6$.

a)\(\dfrac{x}{-4}=\dfrac{-11}{2}\)

x*2=-11*(-4)

x*2=44

x=44/2

x=22

b)\(\dfrac{15-x}{x+9}=\dfrac{3}{5}\)

(15-x)*5=(x+9)*3

$75-5x=3x+27$

$8x=48$

$x=6$.

a)\(\dfrac{x}{-4}=\dfrac{-11}{2}\)

x*2=-11*(-4)

x*2=44

x=44/2

x=22

b)\(\dfrac{15-x}{x+9}=\dfrac{3}{5}\)

(15-x)*5=(x+9)*3

$75-5x=3x+27$

$8x=48$

$x=6$.

a)\(\dfrac{x}{-4}=\dfrac{-11}{2}\)

x*2=-11*(-4)

x*2=44

x=44/2

x=22

b)\(\dfrac{15-x}{x+9}=\dfrac{3}{5}\)

(15-x)*5=(x+9)*3

$75-5x=3x+27$

$8x=48$

$x=6$.

a)\(\dfrac{x}{-4}=\dfrac{-11}{2}\)

x*2=-11*(-4)

x*2=44

x=44/2

x=22

b)\(\dfrac{15-x}{x+9}=\dfrac{3}{5}\)

(15-x)*5=(x+9)*3

$75-5x=3x+27$

$8x=48$

$x=6$.

a)\(\dfrac{x}{-4}=\dfrac{-11}{2}\)

x*2=-11*(-4)

x*2=44

x=44/2

x=22

b)\(\dfrac{15-x}{x+9}=\dfrac{3}{5}\)

(15-x)*5=(x+9)*3

$75-5x=3x+27$

$8x=48$

$x=6$.

a)\(\dfrac{x}{-4}=\dfrac{-11}{2}\)

x*2=-11*(-4)

x*2=44

x=44/2

x=22

b)\(\dfrac{15-x}{x+9}=\dfrac{3}{5}\)

(15-x)*5=(x+9)*3

$75-5x=3x+27$

$8x=48$

$x=6$.